Full answer: In each of 7 through 12, determine the longest interval in which the given

In each of Problems 1 through 6, find the Wronskian of the given pair of functions. e2t , e3t/2

In each of Problems 1 through 6, find the Wronskian of the given pair of functions. cost, sin t

In each of Problems 1 through 6, find the Wronskian of the given pair of functions. e2t , te2t

In each of Problems 1 through 6, find the Wronskian of the given pair of functions. x, xex

In each of Problems 1 through 6, find the Wronskian of the given pair of functions. et sin t, et cost

In each of Problems 1 through 6, find the Wronskian of the given pair of functions. cos2 , 1 + cos 2

In each of Problems 7 through 12, determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the solution. ty + 3y = t, y(1) = 1, y (1) = 2

In each of Problems 7 through 12, determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the solution. (t 1)y 3ty + 4y = sin t, y(2) = 2, y (2) = 1

In each of Problems 7 through 12, determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the solution. t(t 4)y + 3ty + 4y = 2, y(3) = 0, y (3) = 1

In each of Problems 7 through 12, determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the solution. t(t 4)y + 3ty + 4y = 2, y(3) = 0, y (3) = 1

In each of Problems 7 through 12, determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the solution. (x 3)y + xy + (ln |x|)y = 0, y(1) = 0, y (1) = 1

In each of Problems 7 through 12, determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the solution. (x 2)y + y + (x 2)(tan x)y = 0, y(3) = 1, y (3) = 2

Verify that y1(t) = t 2 and y2(t) = t 1 are two solutions of the differential equation t 2y 2y = 0 for t > 0. Then show that y = c1t 2 + c2t 1 is also a solution of this equation for any c1 and c2.

Verify that y1(t) = 1 and y2(t) = t 1/2 are solutions of the differential equation yy + (y )2 = 0 for t > 0. Then show that y = c1 + c2t 1/2 is not, in general, a solution of this equation. Explain why this result does not contradict Theorem 3.2.2

. Show that if y = (t) is a solution of the differential equation y + p(t)y + q(t)y = g(t), where g(t) is not always zero, then y = c(t), where c is any constant other than 1, is not a solution. Explain why this result does not contradict the remark following Theorem 3.2.2

Can y = sin(t 2) be a solution on an interval containing t = 0 of an equation y + p(t)y + q(t)y = 0 with continuous coefficients? Explain your answer.

If the Wronskian W of f and g is 3e4t , and if f(t) = e2t , find g(t).

If the Wronskian W of f and g is t 2et , and if f(t) = t, find g(t)

If W(f, g) is the Wronskian of f and g, and if u = 2f g, v = f + 2g, find the Wronskian W(u, v) of u and v in terms of W(f, g).

If the Wronskian of f and g is t cost sin t, and if u = f + 3g, v = f g, find the Wronskian of u and v.

Assume that y1 and y2 are a fundamental set of solutions of y + p(t)y + q(t)y = 0 and let y3 = a1y1 + a2y2 and y4 = b1y1 + b2y2, where a1, a2,b1, and b2 are any constants. Show that W(y3, y4) = (a1b2 a2b1)W(y1, y2). Are y3 and y4 also a fundamental set of solutions? Why or why not?

In each of Problems 22 and 23, find the fundamental set of solutions specified by Theorem 3.2.5 for the given differential equation and initial point . y + y 2y = 0, t0 = 0

In each of Problems 22 and 23, find the fundamental set of solutions specified by Theorem 3.2.5 for the given differential equation and initial point y + 4y + 3y = 0, t0 = 1

In each of Problems 24 through 27, verify that the functions y1 and y2 are solutions of the given differential equation. Do they constitute a fundamental set of solutions? . y + 4y = 0; y1(t) = cos 2t, y2(t) = sin 2t

In each of Problems 24 through 27, verify that the functions y1 and y2 are solutions of the given differential equation. Do they constitute a fundamental set of solutions? y 2y + y = 0; y1(t) = et , y2(t) = tet

In each of Problems 24 through 27, verify that the functions y1 and y2 are solutions of the given differential equation. Do they constitute a fundamental set of solutions? x2y x(x + 2)y + (x + 2)y = 0, x > 0; y1(x) = x, y2(x) = xex

In each of Problems 24 through 27, verify that the functions y1 and y2 are solutions of the given differential equation. Do they constitute a fundamental set of solutions? . (1 x cot x)y xy + y = 0, 0 < x < ; y1(x) = x, y2(x) = sin x

Consider the equation y y 2y = 0. (a) Show that y1(t) = et and y2(t) = e2t form a fundamental set of solutions. (b) Let y3(t) = 2e2t , y4(t) = y1(t) + 2y2(t), and y5(t) = 2y1(t) 2y3(t). Are y3(t), y4(t), and y5(t) also solutions of the given differential equation? (c) Determine whether each of the following pairs forms a fundamental set of solutions: [y1(t), y3(t)]; [y2(t), y3(t)]; [y1(t), y4(t)]; [y4(t), y5(t)].

In each of Problems 29 through 32, find theWronskian of two solutions of the given differential equation without solving the equation t 2y t(t + 2)y + (t + 2)y = 0

In each of Problems 29 through 32, find theWronskian of two solutions of the given differential equation without solving the equation (cost)y + (sin t)y ty = 0

In each of Problems 29 through 32, find theWronskian of two solutions of the given differential equation without solving the equation x2y + xy + (x2 2)y = 0, Bessels equation

In each of Problems 29 through 32, find theWronskian of two solutions of the given differential equation without solving the equation (1 x2)y 2xy + ( + 1)y = 0, Legendres equation

Show that if p is differentiable and p(t) > 0, then the Wronskian W(t) of two solutions of [p(t)y ]  + q(t)y = 0 is W(t) = c/p(t), where c is a constant.

If the differential equation ty + 2y + tet y = 0 has y1 and y2 as a fundamental set of solutions and if W(y1, y2)(1) = 2, find the value of W(y1, y2)(5).

If the differential equation t 2y 2y + (3 + t)y = 0 has y1 and y2 as a fundamental set of solutions and if W(y1, y2)(2) = 3, find the value of W(y1, y2)(4)

If theWronskian of any two solutions of y + p(t)y + q(t)y = 0 is constant, what does this imply about the coefficients p and q?

If f, g, and h are differentiable functions, show that W(fg, fh) = f 2W(g, h).

In Problems 38 through 40, assume that p and q are continuous and that the functions y1 and y2 are solutions of the differential equation y + p(t)y + q(t)y = 0 on an open interval I. Prove that if y1 and y2 are zero at the same point in I, then they cannot be a fundamental set of solutions on that interval.

In Problems 38 through 40, assume that p and q are continuous and that the functions y1 and y2 are solutions of the differential equation y + p(t)y + q(t)y = 0 on an open interval I. Prove that if y1 and y2 have maxima or minima at the same point in I, then they cannot be a fundamental set of solutions on that interval.

In Problems 38 through 40, assume that p and q are continuous and that the functions y1 and y2 are solutions of the differential equation y + p(t)y + q(t)y = 0 on an open interval I. Prove that if y1 and y2 have a common point of inflection t0 in I, then they cannot be a fundamental set of solutions on I unless both p and q are zero at t0.

Exact Equations. The equation P(x)y + Q(x)y + R(x)y = 0 is said to be exact if it can be written in the form [P(x)y ]  + [f(x)y]  = 0, where f(x) is to be determined in terms of P(x), Q(x), and R(x). The latter equation can be integrated once immediately, resulting in a first order linear equation for y that can be solved as in Section 2.1. By equating the coefficients of the preceding equations and then eliminating f(x), show that a necessary condition for exactness is P(x) Q (x) + R(x) = 0. It can be shown that this is also a sufficient condition.

In each of Problems 42 through 45, use the result of Problem 41 to determine whether the given equation is exact. If it is, then solve the equation. y + xy + y = 0

In each of Problems 42 through 45, use the result of Problem 41 to determine whether the given equation is exact. If it is, then solve the equation. y + 3x2y + xy = 0

In each of Problems 42 through 45, use the result of Problem 41 to determine whether the given equation is exact. If it is, then solve the equation. xy (cos x)y + (sin x)y = 0, x > 0

In each of Problems 42 through 45, use the result of Problem 41 to determine whether the given equation is exact. If it is, then solve the equation. x2y + xy y = 0, x > 0

TheAdjoint Equation. If a second order linear homogeneous equation is not exact,it can be made exact by multiplying by an appropriate integrating factor (x). Thus we require that (x) be such that (x)P(x)y + (x)Q(x)y + (x)R(x)y = 0 can be written in the form [(x)P(x)y ]  + [ f(x)y]  = 0. By equating coefficients in these two equations and eliminating f(x), show that the function must satisfy P + (2P Q) + (P Q + R) = 0. This equation is known as the adjoint of the original equation and is important in the advanced theory of differential equations. In general, the problem of solving the adjoint differential equation is as difficult as that of solving the original equation, so only occasionally is it possible to find an integrating factor for a second order equation.

In each of Problems 47 through 49, use the result of Problem 46 to find the adjoint of the given differential equation. x2y + xy + (x2 2)y = 0, Bessels equation

In each of Problems 47 through 49, use the result of Problem 46 to find the adjoint of the given differential equation. (1 x2)y 2xy + ( + 1)y = 0, Legendres equation

In each of Problems 47 through 49, use the result of Problem 46 to find the adjoint of the given differential equation. y xy = 0, Airys equation

For the second order linear equation P(x)y + Q(x)y + R(x)y = 0, show that the adjoint of the adjoint equation is the original equation.

A second order linear equation P(x)y + Q(x)y + R(x)y = 0 is said to be self-adjoint if its adjoint is the same as the original equation. Show that a necessary condition for this equation to be self-adjoint is that P (x) = Q(x). Determine whether each of the equations in Problems 47 through 49 is self-adjoint.